Question

[Nguyễn Đức Trường]

CM : Nếu \(\frac{a}{b}>1\) thì \(\frac {a+m}{b+m}<\frac {a}b\).

Answer 1

[Riio Riyuko]

\(\frac{a}{b}>1\Rightarrow a>b\)

Ta có :

\(\frac{a+m}{b+m}< \frac{a}{b}\)

<=> \(b\left(a+m\right)< a\left(b+m\right)\)

<=> \(ab+bm< ab+am\)

<=> \(bm< am\)

<=> \(b< a\)  (Đúng do giả thiết cho)

Vậy ......

Answer 2

[Wall HaiAnh]

Ta có: \(\frac{a}{b}=\frac{a\left(b+m\right)}{b\left(b+m\right)}=\frac{ab+am}{b^2+bm}\)

           \(\frac{a+m}{b+m}=\frac{b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+bm}{b^2+bm}\)

\(\Rightarrow\frac{a}{b}>1\Rightarrow a>b\)

\(\Rightarrow ab+am>ab+bm\)

\(\Rightarrow\frac{a+m}{b+m}< \frac{a}{b}\)